Question

Suppose that the inverse demand for San Francisco cable car rides is pequals20minusStartFraction Upper Q Over 1000 EndFraction ​, where p is the price per ride and Q is the number of rides per day. Suppose the objective of San​ Francisco's Municipal Authority​ (the cable car​ operator) is to maximize its revenues LOADING.... What is the​ revenue-maximizing price? The​ revenue-maximizing price is

Answer 1

Answer: The​ revenue-maximizing price is $10.

Explanation:

Given that,

Inverse demand function: P = $20 - \frac{Q}{1,000}$

Where,

P - Price per ride

Q - Number of rides per day

Revenue(R) = P × Q

                   = $20 - \frac{Q}{1,000}$ × Q

                   = $20Q - \frac{Q^{2} }{1,000}$

Differentiating 'R' with respect to Q for calculating Marginal revenue(MR):

MR = $20 - \frac{Q}{500}$

Here, MC = 0

MR = MC

$20 - \frac{Q}{500}$ = 0

Therefore, Q = 10,000

P = $20 - \frac{Q}{1,000}$

  = $20 - \frac{10,000}{1,000}$

  = $10

Hence, the​ revenue-maximizing price is $10.